Thread: Measure Theory -- Integration/Continuity

Hey, I am kind of stumped on this one. Anyone want to offer hints?

If f is in L1(m) and F(x) = integral ( f(t)dt ) from -infinity to x, then F is continuous on R.

I know a couple things, sure:

**F in L1(m) means that {x : f(x) = infinity} is a null set.
**F is integrable (and therefore continuous) is integral(F(x)) < infinity.

Originally Posted by joeyjoejoe

Hey, I am kind of stumped on this one. Anyone want to offer hints?

If f is in L1(m) and F(x) = integral ( f(t)dt ) from -infinity to x, then F is continuous on R.

I know a couple things, sure:

**F in L1(m) means that {x : f(x) = infinity} is a null set.
**F is integrable (and therefore continuous) is integral(F(x)) < infinity.

You should read again your definitions: means that is measurable and integrable with respect to the measure : . And integrable is not linked with continuous.

To show that the function is continuous at , notice that, if , (where the 1 is the indicator function). Then you can use the dominated convergence theorem, when tends to , to show that the previous integral converges to 0.